Import Question JSON

Current Question (ID: 19116)

Question:

$\text{Given below are two statements:}$ $\text{Assertion (A):}$ \text{Two identical balls A and B thrown with the same velocity 'u' at two different angles with horizontal attained the same range R. If A and B reached the maximum height } h_1 \text{ and } h_2 \text{ respectively, then } R = 4\sqrt{h_1 h_2}$ $\text{Reason (R):}$ \text{Product of said heights, } h_1 h_2 = \left( \frac{u^2 \sin^2 \theta}{2g} \right) \cdot \left( \frac{u^2 \cos^2 \theta}{2g} \right)$

Options:

1. $\text{Both (A) and (R) are true and (R) is the correct explanation of (A).}$
2. $\text{Both (A) and (R) are true but (R) is not the correct explanation of (A).}$
3. $\text{(A) is true but (R) is false.}$
4. $\text{(A) is false but (R) is true.}$

Solution:

$\text{The range for a projectile is given by } R = \frac{u^2 \sin 2\theta}{g}.$ $\text{For the same range, } \theta_1 + \theta_2 = 90^\circ.$ $\text{The maximum height is given by } h = \frac{u^2 \sin^2 \theta}{2g}.$ $\text{Thus, } h_1 h_2 = \left( \frac{u^2 \sin^2 \theta_1}{2g} \right) \cdot \left( \frac{u^2 \cos^2 \theta_1}{2g} \right) = \frac{u^4 \sin^2 \theta_1 \cos^2 \theta_1}{4g^2}.$ $\text{Using } \sin 2\theta = 2 \sin \theta \cos \theta, \text{ we get } R = 4\sqrt{h_1 h_2}.$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}